It is not uncommon to find a triple of integers whose squares are in arithmetic progression, e.g.

arithmetic/square.progression.chain New

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Problem

It is not uncommon to find a triple of integers whose squares are in
arithmetic progression, e.g. (41,85,113).  Sometimes two such triples
can be found such that the start of the second is the end of the first,
e.g. (7,13,17) and (17,25,31).  Show that infinitely long chains of
such triples exist.  More precisely, show there is an increasing
"sequence"

           ...,a1,a2,a3,a4,a5,a6,a7,a8,a9,...

of integers such that the squares of each of the triples

     ...,(a1,a2,a3),(a3,a4,a5),(a5,a6,a7),(a7,a8,a9),...

constitutes a three term arithmetic progression.

-- Jim Buddenhagen <jb1556@daditz.sbc.com>

Show solution

Easy.  For each triplet (x,y,z)

        x = 2 (b_n)^2 - 1
        y = 2 (b_n)^2 + 2 b_n + 1
        z = 2 (b_n)^2 + 4 b_n + 1

where   b_(n+1) = (b_n) + 1, and b_1 = 1

Your sequence is

        { 1, 5, 7, 13, 17, 25, 31, ...}

I did not find this example by chance.  Here is my construction.

Suppose x, y, and z are 3 pos. integers such that x^2, y^2, and z^2 are
3 consecutive terms of an A.P.  Obviously, we want to exclude the
trivial case x=y=z.  So suppose x<y<z.  (It is easy to see that no two
of them can be equal either.)  Let x = y - a and z = y + b

Then    x^2 + z^2 = 2 y^2
        (y-a)^2 + (y+b)^2 = 2 y^2
            a^2 + b^2
        y = ---------
             2 (a-b)

2 divides (a^2 + b^2) implies both a and b has the same parity.  Thus
a-b is even.  Say, a - b = 2n

Then    y = (b^2)/(2 n) + b + n

2 divides b^2 implies b is even.  Say b = 2 b_1

Then    y = 2 b_1^2/n + 2 b_1 + n

Since n is arbitrary.  Let's put n=1

Then    y = 2 b_1^2 + 2 b_1 + 1
Consequently,
        x = 2 b_1^2 - 1
        z = 2 b_1^2 + 4 b_1 + 1

Next we want to find (x',y',z') to satisfy the same requirement.  Thus

        x' = 2 b_2^2 - 1
        y' = 2 b_2^2 + 2 b_2 + 1
        z' = 2 b_2^2 + 4 b_2 + 1

In our case x' = z.  So

        2 b_2^2 - 1 = 2 b_1^2 + 4 b_1 + 1
Thus
        b_2 = b_1 + 1

It follows that the sequence given in the above must satisfy your
requirement.

-- Peter Kwok <pkk36438@uxa.cso.uiuc.edu>, UIUC